Question

How do you verify that the hypothesis of mean value theorem is satisfied on the interval \[[2,5]\] and find all the values of c in the given interval that satisfy the conclusion of the theorem for \[f(x)=\dfrac{1}{x-1}\] ?

Answer 1

Hint: Before finding all the values of c in the given interval that satisfy the conclusion of the theorem for the given function, we will check whether the function is continuous on the given closed interval and differentiable on the given open interval. After that in the given interval we will check the values of c which satisfy the mean value theorem.

Complete step by step answer:
In the question the concept of mean value theorem is asked so, let us know what mean value theorem states. The mean value theorem states that if a function is continuous on the closed interval \[[a,b]\] and differentiable on the open interval \[(a,b)\] , then there exists a point \[c\] in the interval \[(a,b)\] such that \[f'(c)\] is equal to functions average rate of change over \[[a,b]\].
According to the mean value theorem \[f'(c)=\dfrac{f(b)-f(a)}{b-a}\].
Now in the given function \[f(x)=\dfrac{1}{x-1}\] we can see that the function is continuous and differentiable at all points except when \[x=1\] . therefore, the given function is continuous on the closed interval \[[2,5]\] and differentiable on the open interval \[(2,5)\]which means we can conclude that there exists at least one point \[c\] in the interval\[(2,5)\] such that \[f'(c)\] is equal to functions average rate of change over \[[2,5]\].
 \[\begin{align}
  & f(5)=\dfrac{1}{4},f(2)=1 \\
 & f'(c)=\dfrac{f(5)-f(2)}{5-2}=\dfrac{\dfrac{1}{4}-1}{3}=\dfrac{\dfrac{-3}{4}}{3}=\dfrac{-1}{4} \\
\end{align}\]
To find the values of c, we will find \[f'(c)\].
\[f'(c)=\dfrac{-1}{{{(c-1)}^{2}}}=-{{(c-1)}^{-2}}\]
Also, we know that \[f'(c)=-\dfrac{1}{4}\]by mean value theorem.
Therefore, equating these two we get,
\[\begin{align}
  & -{{(c-1)}^{-2}}=-\dfrac{1}{4} \\
 & \Rightarrow {{(c-1)}^{2}}=4 \\
 & \Rightarrow c-1=\pm 4 \\
 & \Rightarrow c-1=-4,c-1=4 \\
 & \Rightarrow c=-3,c=5 \\
 & \Rightarrow c=-3,5 \\
\end{align}\]
Hence, the values of c in the given interval that satisfy the conclusion of the theorem for \[f(x)=\dfrac{1}{x-1}\] are \[c=-3,5\].


Note:
While solving the question the first step should be to check whether the function is continuous on the closed interval \[[a,b]\] and differentiable on the open interval \[(a,b)\]. In this question we have used the concept of mean value theorem which is a very important theorem in calculus and is used in many questions. Also do not make mistakes while finding the derivative of the function.